Let us write f(n)(x){\displaystyle f^{(n)}(x)} for the n{\displaystyle n}-th derivative of f(x).{\displaystyle f(x).} We also write f(0)(x)=f(x){\displaystyle f^{(0)}(x)=f(x)} — think of f(x){\displaystyle f(x)} as the "zeroth derivative" of f(x).{\displaystyle f(x).} We thus arrive at the general result f(k)(0)=k!ak,{\displaystyle f^{(k)}(0)=k!\,a_{k},} where the factorialk!{\displaystyle k!} is defined as equal to 1 for k=0{\displaystyle k=0} and k=1{\displaystyle k=1} and as the product of all natural numbers n≤k{\displaystyle n\leq k} for k>1.{\displaystyle k>1.} Expressing the coefficients ak{\displaystyle a_{k}} in terms of the derivatives of f(x){\displaystyle f(x)} at x=0,{\displaystyle x=0,} we obtain

A remarkable result: if you know the value of a well-behaved function f(x){\displaystyle f(x)} and the values of all of its derivatives at the single pointx=0{\displaystyle x=0} then you know f(x){\displaystyle f(x)} at all points x.{\displaystyle x.} Besides, there is nothing special about x=0,{\displaystyle x=0,} so f(x){\displaystyle f(x)} is also determined by its value and the values of its derivatives at any other point x0{\displaystyle x_{0}}: